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Difference between revisions of "2014 AMC 12B Problems/Problem 21"

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[center]<math>a + sin(k) \cdot 1 + cos(k)a = 1</math>
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<cmath>a + sin(k) \cdot 1 + cos(k)a = 1</cmath>
  
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<math>sin(k)a + cos(k) = 1</math>[/center]
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<cmath>sin(k)a + cos(k) = 1</cmath>
 
Solving for <math>a</math> we get:
 
Solving for <math>a</math> we get:
  
  
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[center]<math>a = \frac{1-sin(k)}{cos(k) + 1} = \frac{1 - cos(k)}{sin(k)}</math>
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<cmath>a = \frac{1-sin(k)}{cos(k) + 1} = \frac{1 - cos(k)}{sin(k)}</cmath>
  
 
<math>(1-sin(k)) \cdot sin(k) = (1 - cos(k))\cdot(cos(k) + 1)</math>
 
<math>(1-sin(k)) \cdot sin(k) = (1 - cos(k))\cdot(cos(k) + 1)</math>

Revision as of 10:58, 29 November 2016

Problem 21

In the figure, $ABCD$ is a square of side length $1$. The rectangles $JKHG$ and $EBCF$ are congruent. What is $BE$? [asy] pair A=(1,0), B=(0,0), C=(0,1), D=(1,1), E=(2-sqrt(3),0), F=(2-sqrt(3),1), G=(1,sqrt(3)/2), H=(2.5-sqrt(3),1), J=(.5,0), K=(2-sqrt(3),1-sqrt(3)/2); draw(A--B--C--D--cycle); draw(K--H--G--J--cycle); draw(F--E); label("$A$",A,SE); label("$B$",B,SW); label("$C$",C,NW); label("$D$",D,NE); label("$E$",E,S); label("$F$",F,N); label("$G$",G,E); label("$H$",H,N); label("$J$",J,S); label("$K$",K,W); [/asy] $\textbf{(A) }\frac{1}{2}(\sqrt{6}-2)\qquad\textbf{(B) }\frac{1}{4}\qquad\textbf{(C) }2-\sqrt{3}\qquad\textbf{(D) }\frac{\sqrt{3}}{6}\qquad\textbf{(E) } 1-\frac{\sqrt{2}}{2}$

Solution 1

Draw the altitude from $H$ to $AB$ and call the foot $L$. Then $HL=1$. Consider $HJ$. It is the hypotenuse of both right triangles $\triangle HGJ$ and $\triangle HLJ$, and we know $JG=HL=1$, so we must have $\triangle HGJ\cong\triangle JLH$ by Hypotenuse-Leg congruence. From this congruence we have $LJ=HG=BE$.

Notice that all four triangles in this picture are similar. Also, we have $LA=HD=EJ$. So set $x=LJ=HG=BE$ and $y=LA=HD=EJ$. Now $BE+EJ+JL+LA=2(x+y)=1$. This means $x+y=\frac{1}{2}=BE+EJ=BJ$, so $J$ is the midpoint of $AB$. So $\triangle AJG$, along with all other similar triangles in the picture, is a 30-60-90 triangle, and we have $AG=\sqrt{3} \cdot AJ=\sqrt{3}/2$ and subsequently $GD=\frac{2-\sqrt{3}}{2}=KE$. This means $EJ=\sqrt{3} \cdot KE=\frac{2\sqrt{3}-3}{2}$, which gives $BE=\frac{1}{2}-EJ=\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}$, so the answer is $\textbf{(C)}$.

Solution 2

Let $BE = x$. Let $JA = y$. Because $\angle FKH = \angle EJK = \angle AGJ = \angle DHG$ and $\angle FHK = \angle EKJ = \angle AJG = \angle DGH$, $\triangle KEJ, \triangle JAG, \triangle GDH, \triangle HFK$ are all similar. Using proportions and the pythagorean theorem, we find \[EK = xy\] \[FK = \sqrt{1-y^2}\] \[EJ = x\sqrt{1-y^2}\] Because we know that $BE+EJ+AJ = EK + FK = 1$, we can set up a systems of equations \[x + x\sqrt{1-y^2} + y = 1\] \[xy + \sqrt{1-y^2} = 1\] Solving for $x$ in the second equation, we get \[x= \frac{1-\sqrt{1-y^2}}{y}\] Plugging this into the first equation, we get \[\frac{1-\sqrt{1-y^2}}{y} + (\sqrt{1-y^2})\frac{1-\sqrt{1-y^2}}{y} + y = 1 \implies \frac{2y^2}{y}=1 \implies y=\frac{1}{2}\] Plugging into the previous equation with $x$, we get \[x= 2\left(1-\sqrt{1-\frac{1}{4}}\right) = 2\left(\frac{2-\sqrt{3}}{2} \right) = \boxed{\textbf{(C)}\ 2-\sqrt{3}}\]

Solution 3

Let $BE = x$, $EK = a$, and $EJ = b$. Then $x^2 = a^2 + b^2$ and because $\triangle KEJ \cong \triangle GDH$ and $\triangle KEJ \sim \triangle JAG$, $\frac{GA}{1} = 1 - a = \frac{b}{x}$. Furthermore, the area of the four triangles and the two rectangles sums to 1:

\[1 = 2x + GA\cdot JA + ab\]

\[1 = 2x + (1 - a)(1 - (x + b)) + ab\]

\[1 = 2x + \frac{b}{x}(1 - x - b) + \left(1 - \frac{b}{x}\right)b\]

\[1 = 2x + \frac{b}{x} - b - \frac{b^2}{x} + b - \frac{b^2}{x}\]

\[x = 2x^2 + b - 2b^2\]

\[x - b = 2(x - b)(x + b)\]

\[x + b = \frac{1}{2}\]

\[b = \frac{1}{2} - x\]

\[a = 1 - \frac{b}{x} = 2 - \frac{1}{2x}\]

By the Pythagorean theorem: $x^2 = a^2 + b^2$

\[x^2 = \left(2 - \frac{1}{2x}\right)^2 + \left(\frac{1}{2} - x\right)^2\]

\[x^2 = 4 - \frac{2}{x} + \frac{1}{4x^2} + \frac{1}{4} - x + x^2\]

\[0 = \frac{1}{4x^2} - \frac{2}{x} + \frac{17}{4} - x\]

\[0 = 1 - 8x + 17x^2 - 4x^3.\]

Then by the rational root theorem, this has roots $\frac{1}{4}$, $2 - \sqrt{3}$, and $2 + \sqrt{3}$. The first and last roots are extraneous because they imply $a = 0$ and $x > 1$, respectively, thus $x = \boxed{\textbf{(C)}\ 2-\sqrt{3}}$.

Solution 4

Let $\angle FKH$ = $k$ and $CF$ = $a$. It is shown that all four triangles in the picture are similar. From the square side lengths:


\[a + sin(k) \cdot 1 + cos(k)a = 1\]

\[sin(k)a + cos(k) = 1\] Solving for $a$ we get:


\[a = \frac{1-sin(k)}{cos(k) + 1} = \frac{1 - cos(k)}{sin(k)}\]

$(1-sin(k)) \cdot sin(k) = (1 - cos(k))\cdot(cos(k) + 1)$

$sin(k)-sin(k)^2 = cos(k) + 1 - cos(k)^2 - cos(k)$

$sin(k)-sin(k)^2 = sin(k)^2$

$1-sin(k) = sin(k)$

$sin(k) = \frac{1}{2}, cos(k) = \frac{\sqrt 3}{2}$

$a = \frac{1 - \frac{\sqrt 3}{2}}{\frac{1}{2}} = 2 - \sqrt 3$ [/center]

See also

2014 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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